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<functional>
#include<algorithm>
#include<iterator>
#include<cstring>
#include<cstdio>
using namespace std;
int b[222222];
const int T=262144;
int f[555555];
#define cmin(a,b) ((a)<(b)?:(a)=(b))
#define gmin(a,b) ((a)<(b)?(a):(b))
void C(int l,int x){for(l+=T;l;l>>=1)cmin(f[l],x);}
int Q(int l,int r){int d=233333;for(l+=T-1,r+=T+1;l^r^1;l>>=1,r>>=1)l&1?:cmin(d,f[l^1]),~r&1?:cmin(d,f[r^1]);if(l==0)cmin(d,f[T]);return d;}
struct pii{int l,r;}a[222222];
inline bool operator<(const pii&a,const pii&b){return a.l<b.l?1:a.l==b.l&&a.r<b.r;}
int main()
{
	int n;
	memset(f,0x3f,sizeof(f));
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d",&b[i]);
	int q;
	scanf("%d",&q);
	for(int i=1;i<=q;++i)
		scanf("%d%d",&a[i].l,&a[i].r);
	sort(a+1,a+q+1);
	C(0,0);
	int j=1,ofs=0;
	for(int i=0;i<n;++i)
	{
		for(;a[j].l==i+1;++j)
			C(a[j].r,Q(i,a[j].r));
		if(b[i+1])
			C(i+1,f[i+T]+1);
		else
		{
			++ofs;
			C(i+1,f[i+T]-1);
		}
	}
	printf("%d",f[n+T]+ofs);
	return 0;
}