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 <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
int n,m,i,j,dp[200005],num[200005],x,y,cnt;
vector<int> bi[200005];
int tr[800005];
struct segment_tree
{
    int l,r;
    int sum,add;
    void update(int x)
    {
        sum=max(sum,x);
        add=max(add,x);
    }
}c[400005];
void pushdown(int id)
{
    if (c[id].add)
    {
        c[id*2].update(c[id].add);
        c[id*2+1].update(c[id].add);
        c[id].add=0;
    }
}
void pushup(int id)
{
    c[id].sum=max(c[id*2].sum,c[id*2+1].sum);
}
void build(int id,int l,int r)
{
    c[id].l=l; 
	c[id].r=r;
    c[id].sum=0; 
	c[id].add=0;
    if(l==r) c[id].sum=-1<<30;
    else
    {
        int mid=l+(r-l)/2;
        build(id*2,l,mid);
        build(id*2+1,mid+1,r);
        pushup(id);
    }
}
void update(int id,int le,int ri,int x)
{
    if (le>c[id].r||ri<c[id].l) return;
    if (le<=c[id].l&&c[id].r<=ri) c[id].update(x);
    else
    {
        pushdown(id);
        update(id*2,le,ri,x);
        update(id*2+1,le,ri,x);
        pushup(id);
    }
}
int query(int id,int le,int ri)
{
    if (ri<c[id].l||c[id].r<le) return -1<<30;
    if (le<=c[id].l&&c[id].r<=ri) return c[id].sum;
    pushdown(id);
    return max(query(id*2,le,ri),query(id*2+1,le,ri));
}
int main()
{
	scanf("%d",&n);
	for (i=1;i<=n;i++)
	{
		scanf("%d",num+i);
		if (num[i]==0) cnt++;
	}
	scanf("%d",&m);
	for (i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		bi[x].push_back(y);
	}
	build(1,0,n);
	update(1,0,0,0);
	for (i=1;i<=n;i++)
	{
		for (j=0;j<bi[i].size();j++)
		{
			update(1,bi[i][j],bi[i][j],query(1,i-1,bi[i][j]));
		}
		if (num[i]) update(1,i,i,query(1,i-1,i-1)-1);
			else update(1,i,i,query(1,i-1,i-1)+1);
	}
	cout<<cnt-query(1,n,n);
	return 0;
}