Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <iostream>
#define llint long long
#define mod 1000000007

using namespace std;

llint n, Q;
llint a[3005];
llint dp[3005][3005];
llint inv;

llint modpow(llint a, llint n)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2) % mod;
	}
}

int main(void)
{
	cin >> n >> Q;
	for(int i = 1; i <= n; i++) cin >> a[i];
	inv = modpow(2, mod-2);
	
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			if(a[i] > a[j]) dp[i][j] = 1;
		}
	}
	
	llint l, r;
	for(int q = 0; q < Q; q++){
		cin >> l >> r;
		if(l > r) swap(l, r);
		dp[l][r] = dp[r][l] = (dp[l][r]+dp[r][l])%mod*inv%mod;
		for(int i = 1; i <= n; i++){
			if(i == l || i == r) continue;
			dp[l][i] = dp[r][i] = (dp[l][i]+dp[r][i])%mod*inv%mod;
			dp[i][l] = dp[i][r] = (dp[i][l]+dp[i][r])%mod*inv%mod;
		}
	}
	
	llint ans = 0;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			if(i >= j) continue;
			ans += dp[i][j], ans %= mod;
		}
	}
	ans *= modpow(2, Q), ans %= mod;
	cout << ans << endl;
	
	return 0;
}