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>

using namespace std;
typedef long long ll;
const int MAX = 100010;
ll inv[MAX],fact[MAX],fi[MAX],mod = 1000000007;
void inverse(){
	int i;
	inv[1] = 1;
	for(i=2;i<MAX;i++){
		inv[i] = mod - (mod/i)*inv[mod%i]%mod;
	}
	fact[0] = fi[0] = 1;
	for(i=1;i<MAX;i++){
		fact[i] = fact[i-1]*i%mod;
		fi[i] = (fi[i-1]*inv[i])%mod;
	}
}

ll a[3010],dp[3010][3010] = {};
int main(){
	int i,j,n,q;
	cin >> n >> q;
	for(i=0;i<n;i++){
		cin >> a[i];
	}
	inverse();
	for(i=0;i<n;i++){
		for(j=0;j<n;j++){
			if(a[i]>a[j]) dp[i][j] = 1;
		}
	}
	for(i=0;i<q;i++){
		ll x,y; cin >> x >> y;
		x--; y--;
		for(j=0;j<n;j++){
			if(j==x || j==y) continue;
			ll u = dp[x][j],v = dp[y][j];
			dp[x][j] = (u + v)*inv[2]%mod;
			dp[y][j] = (u + v)*inv[2]%mod;
			u = dp[j][x],v = dp[j][y];
			dp[j][x] = (u + v)*inv[2]%mod;
			dp[j][y] = (u + v)*inv[2]%mod;
		}
		ll u = dp[x][y],v = dp[y][x];
		dp[x][y] = (u + v)*inv[2]%mod;
		dp[y][x] = (u + v)*inv[2]%mod;
	}
	ll ans = 0;
	for(i=0;i<n;i++){
		for(j=i + 1;j<n;j++){
			ans += dp[i][j];
		}
	}
	for(i=0;i<q;i++){
		(ans *= 2) %= mod;
	}
	cout << ans << endl;
}