#include <iostream>
using namespace std; 
 
typedef long long ll;   
 
// Cách tính thứ 3: tam giác Pascal   
 
// - nCk = (n - 1)Ck + (n - 1)C(k - 1)   
// - còn cái ô ở toạ độ (n, k) trong tam giác Pascal có giá trị = nCk   
 
const int N = 1e3 + 5;  
const int MOD = 1e9 + 7;  
 
int C[N][N]; // C[i][j] = iCj    
 
// Chi phí bộ nhớ: O(n^2) 
// Chi phí precompute: O(n^2) 
// Chi phí tính nCk: O(1) 
 
void precompute() {
	for (int i = 0; i < N; i++) {
		C[i][0] = 1;   
		for (int j = 1; j <= i; j++) {
			C[i][j] = C[i - 1][j] + C[i - 1][j - 1];   
			C[i][j] %= MOD; 
		}
	}
}
 
int main() {
	ios::sync_with_stdio(0); cin.tie(0);  
	precompute();   
 
	int t; cin >> t;   
 
	while (t--) {
		int n, k;  
		cin >> n >> k;   
		cout << C[n][k] << '\n'; 
	}
}

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

NQoxMDAwIDEwMDAKNTAwIDQwMAoyMDAgMTAwCjU1IDQwCjEwIDI=

5
1000 1000
500 400
200 100
55 40
10 2