#include <bits/stdc++.h>

#define debug(x)	cerr << #x << ' ' << x << endl

using namespace std;

const int N = 1000005;
const int mod = 1e9 + 7;
const int oo = 1e9;

vector <int> sub[N];
int dp[N], n, q, a[N], S[N];
int t[N * 4];
long long bal[N];

void update(int b, int e, int node, int pos, int val){
	if(b == e){	
		t[node] = max(t[node], val);
		return;
	}
	int mid = (b + e) / 2, l = 2 * node, h = l + 1;
	if(pos <= mid) update(b, mid, l, pos, val);
	else update(mid + 1, e, h, pos, val);
	t[node] = max(t[l], t[h]);
}

int query(int b, int e, int node, int x, int y){
	if(y < b || e < x) return -oo;
	if(b >= x && e <= y) return t[node];
	int mid = (b + e) / 2, l = 2 * node, h = l + 1;
	return max(query(b, mid, l, x, y), query(mid + 1, e, h, x, y));
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%d", &a[i]);
		S[i] = S[i-1] + a[i];
	}
	fill(t, t + N * 4, -oo);
	scanf("%d", &q);
	int x, y;
	for(int i = 1; i <= q; ++i){
		scanf("%d %d", &x, &y);
		sub[x].push_back(y);
	}
	for(int i = 1; i <= n; ++i){
		update(1, n, 1, i, dp[i - 1] - S[i - 1]);
		for(auto kt : sub[i]){
			dp[kt] = max(dp[kt], S[kt] + max(query(1, n, 1, i, kt), dp[i - 1] - S[i - 1]));
			update(1, n, 1, kt, dp[kt] - S[kt]);
		}
		dp[i] = max(dp[i], dp[i - 1] + (a[i] ^ 1));
	}
	cout << n - dp[n] << endl;
	return 0;
}

### Prompt

Please provide a Cpp coded solution 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 <bits/stdc++.h>

#define debug(x)	cerr << #x << ' ' << x << endl

using namespace std;

const int N = 1000005;
const int mod = 1e9 + 7;
const int oo = 1e9;

vector <int> sub[N];
int dp[N], n, q, a[N], S[N];
int t[N * 4];
long long bal[N];

void update(int b, int e, int node, int pos, int val){
	if(b == e){	
		t[node] = max(t[node], val);
		return;
	}
	int mid = (b + e) / 2, l = 2 * node, h = l + 1;
	if(pos <= mid) update(b, mid, l, pos, val);
	else update(mid + 1, e, h, pos, val);
	t[node] = max(t[l], t[h]);
}

int query(int b, int e, int node, int x, int y){
	if(y < b || e < x) return -oo;
	if(b >= x && e <= y) return t[node];
	int mid = (b + e) / 2, l = 2 * node, h = l + 1;
	return max(query(b, mid, l, x, y), query(mid + 1, e, h, x, y));
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%d", &a[i]);
		S[i] = S[i-1] + a[i];
	}
	fill(t, t + N * 4, -oo);
	scanf("%d", &q);
	int x, y;
	for(int i = 1; i <= q; ++i){
		scanf("%d %d", &x, &y);
		sub[x].push_back(y);
	}
	for(int i = 1; i <= n; ++i){
		update(1, n, 1, i, dp[i - 1] - S[i - 1]);
		for(auto kt : sub[i]){
			dp[kt] = max(dp[kt], S[kt] + max(query(1, n, 1, i, kt), dp[i - 1] - S[i - 1]));
			update(1, n, 1, kt, dp[kt] - S[kt]);
		}
		dp[i] = max(dp[i], dp[i - 1] + (a[i] ^ 1));
	}
	cout << n - dp[n] << endl;
	return 0;
}

```