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:
//4182_543_731 is vegetable
#include<cstdio>
using namespace std;
#define N 3005
#define mod 1000000007
#define inv 500000004
int n,dp[N][N],v[N],q,a,b;
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)scanf("%d",&v[i]);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)dp[i][j]=(v[i]<v[j]);
	for(int i=1;i<=q;i++)
	{
		scanf("%d%d",&a,&b);
		int tp=dp[a][b]+dp[b][a];
		for(int j=1;j<=n;j++)
		{
			dp[j][a]=dp[j][b]=1ll*(dp[j][a]+dp[j][b])*inv%mod;
			dp[a][j]=dp[b][j]=1ll*(dp[a][j]+dp[b][j])*inv%mod;
		}
		dp[a][b]=dp[b][a]=1ll*tp*inv%mod;
	}
	int as=0;
	for(int i=2;i<=n;i++)
	for(int j=1;j<i;j++)as=(as+dp[i][j])%mod;
	for(int i=1;i<=q;i++)as=as*2%mod;
	printf("%d\n",as);
}