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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
ll dp[3030][3030];
int d[3030];
int main()
{
	int num, query;
	scanf("%d%d", &num, &query);
	for (int i = 0; i < num; i++)scanf("%d", &d[i]);
	for (int i = 0; i < num; i++)for (int j = 0; j < num; j++)dp[i][j] = d[i] < d[j];
	const ll m=(mod+1)/2;
  	for (int p = 0; p < query; p++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		za--, zb--;
		for (int i = 0; i < num; i++)
		{
			if (i != za&&i != zb)
			{
				dp[za][i] = dp[zb][i] = (dp[za][i] + dp[zb][i])*m % mod;
				dp[i][za] = dp[i][zb] = (dp[i][za] + dp[i][zb])*m % mod;
			}
		}
		dp[za][zb] = dp[zb][za] = (dp[za][zb] + dp[zb][za])*m % mod;
	}
	ll r = 0;
	for (int i = 0; i < num; i++)for (int j = 0; j < i; j++)r = (r + dp[i][j]) % mod;
	for (int i = 0; i < query; i++)r = r * 2 % mod;
	printf("%lld\n", r);
}