#include <bits/stdc++.h>

#define F first
#define S second
#define pb push_back
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int N = 200000 + 10;
const int MOD = 1000000007;
const int LOG = 20;
const int INF = 1000000010;
const int delta = 11353;

int seg[N << 2], DP[N], n, a[N], m, ps[N];
vi R[N];

void shift(int id){
	seg[id << 1] = max(seg[id << 1], seg[id]);
	seg[id << 1 | 1] = max(seg[id << 1 | 1], seg[id]);
}

void mx(int id, int lq, int rq, int x, int l, int r){
	if (rq <= l || r <= lq) return;
	if (lq <= l && r <= rq){
		seg[id] = max(seg[id], x);
		return;
	}
	shift(id);
	int md = (l + r) >> 1;
	mx(id << 1, lq, rq, x, l, md);
	mx(id << 1 | 1, lq, rq, x, md, r);
}

int get(int id, int lq, int rq, int l, int r){
	if (rq <= l || r <= lq)return 0;
	if (lq <= l && r <= rq){
		return seg[id] + ps[l];
	}
	int md = (l + r) >> 1;
	shift(id);
	return(get(id << 1, lq, rq, l, md) + get(id << 1 | 1, lq,rq, md, r));
}

int32_t main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++){
		cin >> a[i];
		ps[i] = a[i] + ps[i - 1];
	}
	for (int i = 0; i < (N << 2); i++) seg[i] = -INF;
	cin >> m;
	for (int i = 1; i <= m; i++){
		int l, r;
		cin >> l >> r;
		R[r].pb(l);
	}
	for (int i = 1; i <= n; i++){
		DP[i] = DP[i - 1] + (a[i] == 0);
		sort(all(R[i]));
		for (auto u:R[i]){
			int ziad = max(DP[u - 1], get(1, u - 1, u, 1, n + 1));	
			//cout << ziad << '\n';
			DP[i] = max(DP[i], ziad + ps[i] - ps[u - 1]);
			mx(1, u, i + 1, ziad - ps[u - 1], 1, n + 1);
		}
	//	cout << i << ' ' << DP[i] << '\n';
	}	
	cout << n - DP[n];




	return 0;
}

### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>

#define F first
#define S second
#define pb push_back
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int N = 200000 + 10;
const int MOD = 1000000007;
const int LOG = 20;
const int INF = 1000000010;
const int delta = 11353;

int seg[N << 2], DP[N], n, a[N], m, ps[N];
vi R[N];

void shift(int id){
	seg[id << 1] = max(seg[id << 1], seg[id]);
	seg[id << 1 | 1] = max(seg[id << 1 | 1], seg[id]);
}

void mx(int id, int lq, int rq, int x, int l, int r){
	if (rq <= l || r <= lq) return;
	if (lq <= l && r <= rq){
		seg[id] = max(seg[id], x);
		return;
	}
	shift(id);
	int md = (l + r) >> 1;
	mx(id << 1, lq, rq, x, l, md);
	mx(id << 1 | 1, lq, rq, x, md, r);
}

int get(int id, int lq, int rq, int l, int r){
	if (rq <= l || r <= lq)return 0;
	if (lq <= l && r <= rq){
		return seg[id] + ps[l];
	}
	int md = (l + r) >> 1;
	shift(id);
	return(get(id << 1, lq, rq, l, md) + get(id << 1 | 1, lq,rq, md, r));
}

int32_t main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++){
		cin >> a[i];
		ps[i] = a[i] + ps[i - 1];
	}
	for (int i = 0; i < (N << 2); i++) seg[i] = -INF;
	cin >> m;
	for (int i = 1; i <= m; i++){
		int l, r;
		cin >> l >> r;
		R[r].pb(l);
	}
	for (int i = 1; i <= n; i++){
		DP[i] = DP[i - 1] + (a[i] == 0);
		sort(all(R[i]));
		for (auto u:R[i]){
			int ziad = max(DP[u - 1], get(1, u - 1, u, 1, n + 1));	
			//cout << ziad << '\n';
			DP[i] = max(DP[i], ziad + ps[i] - ps[u - 1]);
			mx(1, u, i + 1, ziad - ps[u - 1], 1, n + 1);
		}
	//	cout << i << ' ' << DP[i] << '\n';
	}	
	cout << n - DP[n];




	return 0;
}

```