Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int n,q,b[200005],c[200005],dp[200005],seg[1000005];
vector<int>g[200005];
int qry(int ll,int rr,int l,int r,int x)
{
	if(ll>rr||l>r)return 1e9+7;
	if(ll==l&&rr==r)return seg[x];
	int md=(l+r)/2;
	if(rr<=md)return qry(ll,rr,l,md,x*2);
	if(ll>md)return qry(ll,rr,md+1,r,x*2+1);
	return min(qry(ll,md,l,md,x*2),qry(md+1,rr,md+1,r,x*2+1));
}
void upd(int p,int v,int l,int r,int x)
{
//	cout<<p<<" "<<v<<" "<<l<<" "<<r<<" "<<x<<endl;
	if(l==r)
	{
		seg[x]=v;
		return;
	}
	int md=(l+r)/2;
	if(p<=md)upd(p,v,l,md,x*2);else upd(p,v,md+1,r,x*2+1);
	seg[x]=min(seg[x*2],seg[x*2+1]);
}
int main()
{
	scanf("%d",&n);
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		if(!b[i])c[i]=-1;else c[i]=1;
		ans+=(1-b[i]);
	}
	scanf("%d",&q);
	for(int i=1;i<=q;i++){int u,v;scanf("%d%d",&u,&v);g[u].push_back(v);}
	for(int i=1;i<=n;i++)dp[i]=1e9+7,upd(i,dp[i],1,n,1);
	for(int i=1;i<=n;i++)
	{
		for(int k=0;k<g[i].size();k++)
		{
			int j=g[i][k];
			int mi=dp[i-1];
			int mii=mi;
			for(int o=max(i-1,1);o<=j;++o)mii=min(mii,dp[o]);
			mi=min(mi,qry(max(i-1,1),j,1,n,1));
//			cout<<i<<" "<<j<<" "<<mi<<" "<<mii<<endl;
			if(mi<dp[j])
			{
				dp[j]=mi;
				upd(j,mi,1,n,1);
			}
		}
//		dp[i]=min(dp[i],dp[i-1]+c[i]);
		if(dp[i]>dp[i-1]+c[i])
		{
			dp[i]=dp[i-1]+c[i];
			upd(i,dp[i],1,n,1);
		}
//        cout<<i<<" "<<dp[i]<<endl;
//        for(int i=1;i<=n;i++)if(dp[i]>=1e8)printf("INF ");else cout<<dp[i]<<" ";
//        cout<<endl;
//        for(int i=1;i<2*n;i++)if(seg[i]>=1e8)printf("INF ");else cout<<seg[i]<<" ";
//        cout<<endl;
	} 
    printf("%d\n",ans+dp[n]);
    return 0;
}