#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <cassert>
 
#pragma comment(linker, "/STACK:336777216")  
 
using namespace std;
 
const int MAXN = 300000 + 10;
 
int N, Q;
int sb[MAXN], zerob[MAXN];
int b[MAXN];
vector <int> p[MAXN];
int dp[MAXN];
int Tree[MAXN * 4];
 
void Modify(int Node, int L, int R, int Pos, int x)
{
	if (L == R)
	{
		Tree[Node] = x;
		return;
	}
	int Mid = (L + R) / 2;
	if (Pos <= Mid)
		Modify(Node * 2, L, Mid, Pos, x);
	else
		Modify(Node * 2 + 1, Mid + 1, R, Pos, x);
	Tree[Node] = min(Tree[Node * 2], Tree[Node * 2 + 1]);
}
 
int Query(int Node, int L, int R, int l, int r)
{
	if (L == l && r == R)
		return Tree[Node];
	int Mid = (L + R) / 2;
	if (r <= Mid)
		return Query(Node * 2, L, Mid, l, r);
	else if (l > Mid)
		return Query(Node * 2 + 1, Mid + 1, R, l, r);
	else
		return min(Query(Node * 2, L, Mid, l, Mid), Query(Node * 2 + 1, Mid + 1, R, Mid + 1, r));
}
 
void Work()
{
	scanf("%d", &N);
	for (int i = 1; i <= N; i ++)
	{
		scanf("%d", &b[i]);
		zerob[i] = zerob[i - 1] + (1 - b[i]);
	}
 
	scanf("%d", &Q);
	for (int i = 0; i < Q; i ++)
	{
		int l, r;
		scanf("%d%d", &l, &r);
		p[l].push_back(r);
	}
	
	dp[0] = 0;
	memset(Tree, 1, sizeof(Tree));
	for (int i = 1; i <= N; i ++)
		dp[i] = 10000000;
 
	for (int i = 1; i <= N; i ++)
	{
		Modify(1, 1, N, i, dp[i - 1] - zerob[i - 1]);
 
		for (int j: p[i]) // [i, j]
		{
			// dp[j] <- dp[k - 1] + zerob[j] - zerob[k - 1];
			int tmp = Query(1, 1, N, i, j);
			dp[j] = min(dp[j], zerob[j] + tmp);
			if (j != N)
				Modify(1, 1, N, j + 1, dp[j] - zerob[j]);
		}
		dp[i] = min(dp[i], dp[i - 1] + (b[i] != 0));
	}
	printf("%d\n", dp[N]);
}
 
int main()
{
	Work();
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <cassert>
 
#pragma comment(linker, "/STACK:336777216")  
 
using namespace std;
 
const int MAXN = 300000 + 10;
 
int N, Q;
int sb[MAXN], zerob[MAXN];
int b[MAXN];
vector <int> p[MAXN];
int dp[MAXN];
int Tree[MAXN * 4];
 
void Modify(int Node, int L, int R, int Pos, int x)
{
	if (L == R)
	{
		Tree[Node] = x;
		return;
	}
	int Mid = (L + R) / 2;
	if (Pos <= Mid)
		Modify(Node * 2, L, Mid, Pos, x);
	else
		Modify(Node * 2 + 1, Mid + 1, R, Pos, x);
	Tree[Node] = min(Tree[Node * 2], Tree[Node * 2 + 1]);
}
 
int Query(int Node, int L, int R, int l, int r)
{
	if (L == l && r == R)
		return Tree[Node];
	int Mid = (L + R) / 2;
	if (r <= Mid)
		return Query(Node * 2, L, Mid, l, r);
	else if (l > Mid)
		return Query(Node * 2 + 1, Mid + 1, R, l, r);
	else
		return min(Query(Node * 2, L, Mid, l, Mid), Query(Node * 2 + 1, Mid + 1, R, Mid + 1, r));
}
 
void Work()
{
	scanf("%d", &N);
	for (int i = 1; i <= N; i ++)
	{
		scanf("%d", &b[i]);
		zerob[i] = zerob[i - 1] + (1 - b[i]);
	}
 
	scanf("%d", &Q);
	for (int i = 0; i < Q; i ++)
	{
		int l, r;
		scanf("%d%d", &l, &r);
		p[l].push_back(r);
	}
	
	dp[0] = 0;
	memset(Tree, 1, sizeof(Tree));
	for (int i = 1; i <= N; i ++)
		dp[i] = 10000000;
 
	for (int i = 1; i <= N; i ++)
	{
		Modify(1, 1, N, i, dp[i - 1] - zerob[i - 1]);
 
		for (int j: p[i]) // [i, j]
		{
			// dp[j] <- dp[k - 1] + zerob[j] - zerob[k - 1];
			int tmp = Query(1, 1, N, i, j);
			dp[j] = min(dp[j], zerob[j] + tmp);
			if (j != N)
				Modify(1, 1, N, j + 1, dp[j] - zerob[j]);
		}
		dp[i] = min(dp[i], dp[i - 1] + (b[i] != 0));
	}
	printf("%d\n", dp[N]);
}
 
int main()
{
	Work();
	return 0;
}
```