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<cstdio>
#define llong long long
using namespace std;

const int N = 3000;
const int P = 1e9+7;
const int INV2 = 5e8+4;
llong a[N+3];
llong dp[N+3][N+3];
int n,q;

llong quickpow(llong x,llong y)
{
	llong cur = x,ret = 1ll;
	for(int i=0; y; i++)
	{
		if(y&(1ll<<i)) {y-=(1ll<<i); ret = ret*cur%P;}
		cur = cur*cur%P;
	}
	return ret;
}

int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1; i<=n; i++)
	{
		scanf("%d",&a[i]);
	}
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<i; j++) dp[i][j] = a[j]<a[i] ? 1 : 0;
		for(int j=i+1; j<=n; j++) dp[i][j] = a[i]>a[j] ? 1 : 0;
	}
	for(int i=1; i<=q; i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		dp[x][y] = dp[y][x] = (dp[x][y]+dp[y][x])*INV2%P;
		for(int j=1; j<=n; j++)
		{
			if(j!=x && j!=y)
			{
				dp[j][x] = dp[j][y] = (dp[j][x]+dp[j][y])*INV2%P;
				dp[x][j] = dp[y][j] = (dp[x][j]+dp[y][j])*INV2%P;
			}
		}
	}
	llong ans = 0ll;
	for(int i=1; i<=n; i++)
	{
		for(int j=i+1; j<=n; j++)
		{
			ans = (ans+dp[i][j])%P;
		}
	}
	ans = ans*quickpow(2ll,q)%P;
	printf("%lld\n",ans);
	return 0;
}