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>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

int n,n2=1,q,b[200010],x,y,add=0,cost[200010],dp[200010],dat[800010];
vector <int> v[200010];

void upd(int k,int val)
{
	dat[k]=min(dat[k],val);
	while(k>0)
	{
		k=(k-1)/2;
		dat[k]=min(dat[k*2+1],dat[k*2+2]);
	}
}

int query(int k,int lb,int ub,int tlb,int tub)
{
	if(ub<tlb||tub<lb) return 1e9;
	if(tlb<=lb&&ub<=tub) return dat[k];
	return min(query(k*2+1,lb,(lb+ub)/2,tlb,tub),query(k*2+2,(lb+ub)/2+1,ub,tlb,tub));
}

int main()
{
	cin>>n;
	repn(i,n)
	{
		scanf("%d",&b[i]);
		cost[i]=(b[i]==1 ? 1:-1);
		add+=(b[i]==0);
	}
	cin>>q;
	rep(i,q)
	{
		scanf("%d%d",&x,&y);
		v[x].pb(y);
	}
	repn(i,200005) dp[i]=1e9;
	while(n2<n) n2*=2;
	rep(i,800005) dat[i]=1e9;
	repn(i,n)
	{
		rep(j,v[i].size())
		{
			int ne=min(dp[i-1],query(0,0,n2-1,i,v[i][j]-1));
			if(ne<dp[v[i][j]])
			{
				dp[v[i][j]]=ne;
				upd(v[i][j]+n2-1,dp[v[i][j]]);
			}
		}
		dp[i]=min(dp[i],dp[i-1]+cost[i]);
	}
	cout<<dp[n]+add<<endl;
	return 0;
}