#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; 
struct interval
{
	int l, r; 
	inline bool operator <(const interval &a) const
	{
		if (l == a.l)
			return r > a.r; 
		return l < a.l; 
	}
} arr[200005]; 
int mn[400005], b[200005], pre[200005], n; 
void modify(int pos, int x)
{
	pos += n + 1; 
	for (mn[pos] = min(mn[pos], x); pos; pos >>= 1)
		mn[pos >> 1] = min(mn[pos], mn[pos ^ 1]); 
}
int query(int l, int r)
{
	r++; 
	int ans = 1e9; 
	for (l += n + 1, r += n + 1; l < r; l >>= 1, r >>= 1)
	{
		if (l & 1)
			ans = min(ans, mn[l++]); 
		if (r & 1)
			ans = min(ans, mn[--r]); 
	}
	return ans; 
}
int main()
{
	// freopen("ARC085-F.in", "r", stdin); 
	scanf("%d", &n); 
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", b + i); 
		pre[i] = pre[i - 1] + !b[i]; 
	}
	int q; 
	scanf("%d", &q); 
	for (int i = 0; i < q; i++)
		scanf("%d%d", &arr[i].l, &arr[i].r); 
	sort(arr, arr + q); 
	memset(mn, 0x3f, sizeof(mn)); 
	int idx = 0, ans = 0; 
	for (int i = 1; i <= n; i++)
	{
		while (idx < q && arr[idx].l == i)
		{
			int res = pre[arr[idx].r] + min(query(i, arr[idx].r), ans - pre[i - 1]); 
			modify(arr[idx].r, res - pre[arr[idx].r]); 
			idx++; 
		}
		ans = min(ans + b[i], query(i, i) + pre[i]); 
	}
	printf("%d\n", ans);
	return 0; 
}

### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; 
struct interval
{
	int l, r; 
	inline bool operator <(const interval &a) const
	{
		if (l == a.l)
			return r > a.r; 
		return l < a.l; 
	}
} arr[200005]; 
int mn[400005], b[200005], pre[200005], n; 
void modify(int pos, int x)
{
	pos += n + 1; 
	for (mn[pos] = min(mn[pos], x); pos; pos >>= 1)
		mn[pos >> 1] = min(mn[pos], mn[pos ^ 1]); 
}
int query(int l, int r)
{
	r++; 
	int ans = 1e9; 
	for (l += n + 1, r += n + 1; l < r; l >>= 1, r >>= 1)
	{
		if (l & 1)
			ans = min(ans, mn[l++]); 
		if (r & 1)
			ans = min(ans, mn[--r]); 
	}
	return ans; 
}
int main()
{
	// freopen("ARC085-F.in", "r", stdin); 
	scanf("%d", &n); 
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", b + i); 
		pre[i] = pre[i - 1] + !b[i]; 
	}
	int q; 
	scanf("%d", &q); 
	for (int i = 0; i < q; i++)
		scanf("%d%d", &arr[i].l, &arr[i].r); 
	sort(arr, arr + q); 
	memset(mn, 0x3f, sizeof(mn)); 
	int idx = 0, ans = 0; 
	for (int i = 1; i <= n; i++)
	{
		while (idx < q && arr[idx].l == i)
		{
			int res = pre[arr[idx].r] + min(query(i, arr[idx].r), ans - pre[i - 1]); 
			modify(arr[idx].r, res - pre[arr[idx].r]); 
			idx++; 
		}
		ans = min(ans + b[i], query(i, i) + pre[i]); 
	}
	printf("%d\n", ans);
	return 0; 
}

```