One-X
In this sad world full of imperfections, ugly segment trees exist.A segment tree is a tree where each node represents a segment and has its number. A segment tree for an array of $$$n$$$ elements can be built in a recursive manner. Let's say function $$$\operatorname{build}(v,l,r)$$$ builds the segment tree rooted in the node with number $$$v$$$ and it corresponds to the segment $$$[l,r]$$$.Now let's define $$$\operatorname{build}(v,l,r)$$$:   If $$$l=r$$$, this node $$$v$$$ is a leaf so we stop adding more edges  Else, we add the edges $$$(v, 2v)$$$ and $$$(v, 2v+1)$$$. Let $$$m=\lfloor \frac{l+r}{2} \rfloor$$$. Then we call $$$\operatorname{build}(2v,l,m)$$$ and $$$\operatorname{build}(2v+1,m+1,r)$$$. So, the whole tree is built by calling $$$\operatorname{build}(1,1,n)$$$.Now Ibti will construct a segment tree for an array with $$$n$$$ elements. He wants to find the sum of $$$\operatorname{lca}^\dagger(S)$$$, where $$$S$$$ is a non-empty subset of leaves. Notice that there are exactly $$$2^n - 1$$$ possible subsets. Since this sum can be very large, output it modulo $$$998\,244\,353$$$.$$$^\dagger\operatorname{lca}(S)$$$ is the number of the least common ancestor for the nodes that are in $$$S$$$.
  In the first test case:Let's look at all subsets of leaves.  $$$\operatorname{lca}(\{2\})=2$$$;  $$$\operatorname{lca}(\{3\})=3$$$;  $$$\operatorname{lca}(\{2,3\})=1$$$. Thus, the answer is $$$2+3+1=6$$$.In the second test case:Let's look at all subsets of leaves.  $$$\operatorname{lca}(\{4\})=4$$$;  $$$\operatorname{lca}(\{5\})=5$$$;  $$$\operatorname{lca}(\{3\})=3$$$;  $$$\operatorname{lca}(\{4,5\})=2$$$;  $$$\operatorname{lca}(\{4,3\})=1$$$;  $$$\operatorname{lca}(\{5,3\})=1$$$;  $$$\operatorname{lca}(\{4,5,3\})=1$$$; Thus, the answer is $$$4+5+3+2+1+1+1=17$$$.
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^{18}$$$) — the length of the array for which the segment tree is built.
For each test case, output a single integer — the required sum modulo $$$998\,244\,353$$$.
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()
dp = {1: (1, 0)}
mo = 998244353

def solve(n):
    if n in dp:
        return dp[n]
    (l, r) = ((n + 1) // 2, n - (n + 1) // 2)
    a = (pow(2, l, mo) - 1) * (pow(2, r, mo) - 1) % mo
    (la, lb) = solve(l)
    a += 2 * la % mo
    b = lb
    (ra, rb) = solve(r)
    a += 2 * ra % mo
    b += ra + rb
    dp[n] = (a % mo, b % mo)
    return dp[n]
for _ in range(int(input())):
    n = readInt()
    x = solve(n)
    print((x[0] + x[1]) % mo)
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()
dp = {1: (1, 0)}
mo = 998244353

def solve(n):
    if n in dp:
        return dp[n]
    (l, r) = ((n + 1) // 2, n - (n + 1) // 2)
    a = (pow(2, l, mo) - 1) * (pow(2, r, mo) - 1) % mo
    (la, lb) = solve(l)
    a += 2 * la % mo
    b = lb
    (ra, rb) = solve(r)
    a += 2 * ra % mo
    b += ra + rb
    dp[n] = (a % mo, b % mo)
    return dp[n]
for _ in range(int(input())):
    n = readInt()
    x = solve(n)
    print((x[0] + x[1]) % mo)
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()
dp = {1: (1, 0)}
mo = 998244353

def solve(n):
    if n in dp:
        return dp[n]
    (l, r) = ((n + 1) // 2, n - (n + 1) // 2)
    a = (pow(2, l, mo) - 1) * (pow(2, r, mo) - 1) % mo
    (la, lb) = solve(l)
    a += 2 * la % mo
    b = lb
    (ra, rb) = solve(r)
    a += 2 * ra % mo
    b += ra + rb
    dp[n] = (a % mo, b % mo)
    return dp[n]
for _ in range(int(input())):
    n = readInt()
    x = solve(n)
    print((x[0] + x[1]) % mo)