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 "bits/stdc++.h"
using namespace std;
#define fi first
#define se second
#define ll long long
#define dbg(v) cerr<<#v<<" = "<<v<<'\n'
#define vi vector<int>
#define vl vector <ll>
#define pii pair<int,int>
#define mp make_pair
#define db long double
#define pb push_back
#define all(s) s.begin(),s.end()
template < class T > T smin(T &a,T b) {if (a > b) a = b;return a;}
template < class T > T smax(T &a,T b) {if (a < b) a = b;return a;}
const int N = (int)(1e6) + 5;
const int oo = 1e8;
int t[N];
int s[N];
int g(int l,int r)
{
    return s[r] - s[l - 1];
}
void update(int l,int r,int pos,int v,int node)
{
    if (l == r)
        t[node] = v;
    else
    {
        int m = (l + r) / 2;
        if (pos <= m)
            update(l,m,pos,v,node * 2);
        else
            update(m+1,r,pos,v,node * 2 + 1);
        t[node] = min(t[node * 2],t[node * 2 + 1]);
    }
}
int query(int l,int r,int l1,int r1,int node)
{
    if (l1 <= l && r <= r1)
        return t[node];
    else
    {
        int m = (l + r) / 2;
        int ans = oo;
        if (l1 <= m)
            smin(ans,query(l,m,l1,r1,node * 2));
        if (m+1<=r1)
            smin(ans,query(m+1,r,l1,r1,node * 2 + 1));
        return ans;
    }
}
void build(int l,int r,int node)
{
    t[node] = oo;
    if (l == r) return;
    int m = (l + r) / 2;
    build(l,m,node * 2);
    build(m+1,r,node * 2 + 1);
}
vi w[N];
int dp[N];
int main(void)
{
    int n;
    ios_base :: sync_with_stdio(0);
    cin>>n;
    for (int i = 1;i <= n;++i)
        cin>>s[i],s[i] += s[i - 1];
    int m;
    cin>>m;
    while (m --)
    {
        int l,r;
        cin>>l>>r;
        w[l].pb(r);
    }
    build(1,n,1);
    for (int i = 1;i <= n;++i)
        dp[i] = oo;
    for (int i = 1;i <= n;++i)
    {
        smin(dp[i],dp[i - 1] + g(i,i));
        for (auto it : w[i])
        {
            int omg = min(query(1,n,i,it,1),dp[i - 1] + s[i - 1] - (i - 1));
            update(1,n,it,omg,1);
            smin(dp[it],omg + it - s[it]);
        }
    }
    cout << dp[n] << '\n';
    return 0;
}