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 <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>

using namespace std;

const int mod = 1000000007;

long long qpow(long long base,int tms)
{
	long long tmp = 1;
	while(tms)
	{
		if(tms & 1) tmp = tmp * base % mod;
		base = base * base % mod;
		tms >>= 1;
	}
	return tmp;
}

long long inv2 = qpow(2,mod - 2);
long long dp[3050][3050];
int N,Q;
int seq[3050];

int main()
{
	scanf("%d%d",&N,&Q);
	for(int i = 1;i <= N; ++ i)
		scanf("%d",&seq[i]);
	for(int i = 1;i <= N; ++ i)
		for(int j = 1;j <= N; ++ j)
			dp[i][j] = seq[i] > seq[j];
	long long make_me_burn = qpow(2,Q);
	while(Q --)
	{
		int x,y; scanf("%d%d",&x,&y);
		dp[x][y] = dp[y][x] = (dp[x][y] + dp[y][x]) * inv2 % mod;
		for(int i = 1;i <= N; ++ i)
			if(i != x && i != y)
				dp[i][x] = dp[i][y] = (dp[i][x] + dp[i][y]) * inv2 % mod,
				dp[x][i] = dp[y][i] = (dp[x][i] + dp[y][i]) * inv2 % mod;
	}
	long long ans = 0;
	for(int i = 1;i <= N; ++ i)
		for(int j = i + 1;j <= N; ++ j)
			ans = ans + dp[i][j] * make_me_burn % mod,ans %= mod;
	printf("%lld\n",ans);
}