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>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,q,i,x,y,j,dp[3005][3005],p2[3005],ans=0,st[3005][3005],a[3005],u,v;
const ll mod=1e9+7;
int main()
{
	p2[0]=1;for(i=1;i<=3000;i++)p2[i]=p2[i-1]*2%mod;
	scanf("%lld%lld",&n,&q);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(i=1;i<=n;i++)
	{
		for(j=i+1;j<=n;j++)
		{
			if(a[i]==a[j])continue;
			if(a[i]>a[j])dp[j][i]++;
			else dp[i][j]++;
		}
	}
	for(i=1;i<=q;i++)
	{
		scanf("%lld%lld",&x,&y);
		for(j=1;j<=n;j++)
		{
			if(j==x||j==y)continue;
			u=dp[x][j]=dp[x][j]*p2[i-1-st[x][j]]%mod;st[x][j]=i;
			v=dp[y][j]=dp[y][j]*p2[i-1-st[y][j]]%mod;st[y][j]=i;
			dp[x][j]=(dp[x][j]+v)%mod;
			dp[y][j]=(dp[y][j]+u)%mod;
			u=dp[j][x]=dp[j][x]*p2[i-1-st[j][x]]%mod;st[j][x]=i;
			v=dp[j][y]=dp[j][y]*p2[i-1-st[j][y]]%mod;st[j][y]=i;
			dp[j][x]=(dp[j][x]+v)%mod;
			dp[j][y]=(dp[j][y]+u)%mod;
		}
		u=dp[x][y]=dp[x][y]*p2[i-1-st[x][y]]%mod;st[x][y]=i;
		v=dp[y][x]=dp[y][x]*p2[i-1-st[y][x]]%mod;st[y][x]=i;
		dp[x][y]=(dp[x][y]+v)%mod;
		dp[y][x]=(dp[y][x]+u)%mod;
	}
	for(i=1;i<=n;i++)for(j=1;j<i;j++)ans=(ans+dp[i][j]*p2[q-st[i][j]])%mod;
	printf("%lld\n",ans);
	return 0;
}