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 <bits/stdc++.h>
using namespace std;
typedef long long LL;

const LL mod = 1e9 + 7;
const int N = 3010;

inline LL power(LL a, LL n, LL mod)
{
	LL Ans = 1;
	while (n)
	{
		if (n & 1) Ans = Ans * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return Ans;
}

int a[N], X[N], Y[N];
int n, q;

LL dp[N][N];

int main()
{
	LL inv2 = power(2, mod-2, mod);
	
	cin >> n >> q;
	for (int i = 1; i <= n; i++) scanf("%d", a + i);
	for (int i = 1; i <= q; i++) scanf("%d%d", X + i, Y + i);
	
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) dp[i][j] = a[i] < a[j] ? 1 : 0;
	
	int cur = 0;
	for (int i = 1; i <= q; i++)
	{
		cur ^= 1;
		int a = X[i], b = Y[i];
		dp[a][b] = dp[b][a] = 1LL * inv2 * (dp[a][b] + dp[b][a]) % mod;
		for (int j = 1; j <= n; j++)
			if (j != a && j != b)
			{
				dp[j][a] = dp[j][b] = 1LL * inv2 * (dp[j][a] + dp[j][b]) % mod;
				dp[a][j] = dp[b][j] = 1LL * inv2 * (dp[a][j] + dp[b][j]) % mod;
			}
	}
	
	LL Ans = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j < i; j++) Ans = (Ans + dp[i][j]) % mod;
	cout << Ans * power(2, q, mod) % mod << endl;
	
	return 0;
}