https://codeforces.com/contest/1556/problem/E
令 \(c[i] = a[i]-b[i]\)，将对 \(a\) 的操作看作 \(+1\)，对 \(b\) 的操作看作 \(-1\)，奇数位加，偶数位减，可以想到括号序列，\(c[i]<0\)，即为连续的左括号，\(c[i]>0\)，即为连续的右括号，问题转化为每次可以删除括号序列中若干 \(()\)，问最少需要删除几次
首先判断括号序列是否有解，维护 \(c[i]\) 的前缀和，要求 \(l\) 到 \(r\) 的每个位置右括号均多于或等于左括号的数量，维护前缀和的区间最大值即可
如果有解，左括号累计最多的位置的左括号数量即为答案，因为此时每个左括号都不能同时被删除，且所有删除操作中都包含其中一个左括号
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200010;
const ll INF = 1e18+7;
int n, q;
ll a[maxn], b[maxn], c[maxn], sum[maxn];
struct Node{
ll mi, mx;
}t[maxn<<2];
void pushup(int i){
t[i].mi = min(t[i<<1].mi, t[i<<1|1].mi);
t[i].mx = max(t[i<<1].mx, t[i<<1|1].mx);
}
void build(int i, int l, int r){
if(l == r) {
t[i].mi = t[i].mx = sum[l];
return;
}
int mid = (l + r) >> 1;
build(i<<1, l, mid); build(i<<1|1, mid+1, r);
pushup(i);
}
ll qmi(int i, int l, int r, int x, int y){
if(x <= l && r <= y){
return t[i].mi;
}
int mid = (l + r) >> 1;
ll res = INF;
if(x <= mid) res = min(res, qmi(i<<1, l, mid, x, y));
if(y > mid) res = min(res, qmi(i<<1|1, mid+1, r, x, y));
return res;
}
ll qmx(int i, int l, int r, int x, int y){
if(x <= l && r <= y){
return t[i].mx;
}
int mid = (l + r) >> 1;
ll res = -INF;
if(x <= mid) res = max(res, qmx(i<<1, l, mid, x, y));
if(y > mid) res = max(res, qmx(i<<1|1, mid+1, r, x, y));
return res;
}
ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }
int main(){
n = read(), q = read();
for(int i = 1 ; i <= n ; ++i) a[i] = read();
for(int i = 1 ; i <= n ; ++i) b[i] = read();
for(int i = 1 ; i <= n ; ++i) c[i] = a[i] - b[i];
for(int i = 1 ; i <= n ; ++i) sum[i] = sum[i-1] + c[i];
// for(int i = 1 ; i <= n ; ++i) printf("%d ", c[i]); printf("\n");
// for(int i = 1 ; i <= n ; ++i) printf("%d ", sum[i]); printf("\n");
build(1, 1, n);
int l, r;
for(int i = 1 ; i <= q ; ++i){
l = read(), r = read();
if(qmx(1, 1, n, l, r) <= sum[l-1] && qmx(1, 1, n, r, r) == sum[l-1]){
printf("%lld\n", -(qmi(1, 1, n, l, r)-sum[l-1]));
} else{
printf("-1\n");
}
}
return 0;
}