# #925753 (PyPy3) No.2529 Treasure Hunter - yukicoder
# https://yukicoder.me/submissions/925753
import sys
from itertools import permutations
from heapq import heappop,heappush
from collections import deque
import random
import bisect

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

def cmb(n, r, mod):
    if ( r<0 or r>n ):
        return 0
    return (g1[n] * g2[r] % mod) * g2[n-r] % mod


mod = 998244353
N = 2*10**5
g1 = [1]*(N+1)
g2 = [1]*(N+1)
inverse = [1]*(N+1)

for i in range( 2, N + 1 ):
    g1[i]=( ( g1[i-1] * i ) % mod )
    inverse[i]=( ( -inverse[mod % i] * (mod//i) ) % mod )
    g2[i]=( (g2[i-1] * inverse[i]) % mod )
inverse[0]=0

def solve(N,M):
    if N <= 3:
        dp = [1,0]
        for _ in range(M):
            dp = [(dp[0]+dp[1]) % mod,(dp[0]*N+dp[1]*(N-1)) % mod]
        return (dp[0] + dp[1]) % mod
    
    dp = [1,0,0]
    n = N
    for _ in range(M):
    	# 解説 No.2529 Treasure Hunter - yukicoder --- 解说第2529期 宝藏猎人 - yukicoder
		# https://yukicoder.me/problems/no/2529/editorial
		# 任选 2 个，去掉相邻的情况(n 个)
        a = n*(n-1)//2-n
        # n-1 个任选 2 个，去掉相邻的情况 n-2*j
        b = (n-1)*(n-2)//2-(n-2)
        c = (n-2)*(n-3)//2-(n-4)
        dp = [
            (dp[0]+dp[1]+dp[2]) % mod,
            (dp[0]*N + dp[1]*(N-1) + dp[2]*(N-2)) % mod,
            (dp[0]*a % mod + dp[1] * b % mod + dp[2] * c % mod) % mod
        ]
    
    return sum(dp) % mod

for _ in range(int(input())):
    N,M = mi()
    print(solve(N,M))