/*
首先對於年份進行離散化
然後我們的查詢有兩種，一個是查詢區間最大值，一個是查詢區間是否連續
更改操作有一種 單點修改
我的方法是用線段樹維護最大值，用並查集維護是否連通
抱歉沒有單點修改操作 QWQ 那麽我們在離散化的時候     添加虛節點表示是否會有未知的
用一個last數組記錄最遠的連續的位置
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define M 200010
#define
 
 lson l, mid , now << 1
#define rson mid + 1, r, now << 1 | 1
using namespace std;

int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == ‘-‘) f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - ‘0‘;
    return nm * f;
}
 
int t[M << 2];

int y[M], lst[M], note[M], n, tmp, m, cnt;

void modify(int l, int r, int now, int k, int ver) {
    if(l > k || r < k) return ;
    if(l == r) {
        t[now] = ver;
        return;
    }
    int mid = (l + r) >> 1;
    modify(lson, k, ver);
    modify(rson, k, ver);
    t[now] 
 
= max(t[now << 1], t[now << 1 | 1]);
}

int query(int l, int r, int now, int ln, int rn) {
    if(l > rn || r < ln) return 0;
    if(l >= ln && r <= rn) return t[now];
    int mid = (l + r) >> 1;
    return max(query(lson, ln, rn), query(rson, ln, rn));
}
int main() {
    n = read();
    for(int i = 1; i <= n; i++) {
        y[i] = read();
        if(y[i] == y[i - 1] + 1) {
            note[++cnt] = y[i];
            lst[cnt] = lst[cnt - 1];
        } else {
            note[++cnt] = y[i] - 1;
            lst[cnt] = cnt;
            modify(1, 200000, 1, cnt, 0);
            note[++cnt] = y[i];
            lst[cnt] = cnt;
        }
        modify(1, 200000, 1, cnt, read());
    }
    cnt++;
    lst[cnt] = cnt;
    note[cnt] = 0x3e3e3e3e;
    modify(1, 200000, 1, cnt, 0);
    m = read();
    while(m--) {
        int yn = read(), xn = read();
        yn = lower_bound(note + 1, note + cnt + 1, yn) - note;
        xn = lower_bound(note + 1, note + cnt + 1, xn) - note;
        int ox = query(1, 200000, 1, xn, xn), oy = query(1, 200000, 1, yn, yn), om = query(1, 200000, 1, yn + 1, xn - 1);
        bool fx = (ox > 0), fy = (oy > 0);
        if((fx && om >= ox) || (fy && om >= oy) || (fx && fy && oy <= ox)){puts("false"); continue;}
        if(lst[xn] > yn || !fx || !fy){puts("maybe"); continue;}
        puts("true");
    }
return 0;
}

[SCOI2007]降雨量