Problem: You are given an integer N. Consider an infinite N-ary tree as shown below:

<image>

Figure: an infinite N-ary tree for the case N = 3

As shown in the figure, each vertex is indexed with a unique positive integer, and for every positive integer there is a vertex indexed with it. The root of the tree has the index 1. For the remaining vertices, vertices in the upper row have smaller indices than those in the lower row. Among the vertices in the same row, a vertex that is more to the left has a smaller index.

Regarding this tree, process Q queries. The i-th query is as follows:

* Find the index of the lowest common ancestor (see Notes) of Vertex v_i and w_i.

Constraints

* 1 ≤ N ≤ 10^9
* 1 ≤ Q ≤ 10^5
* 1 ≤ v_i < w_i ≤ 10^9

Input

Input is given from Standard Input in the following format:


N Q
v_1 w_1
:
v_Q w_Q


Output

Print Q lines. The i-th line (1 ≤ i ≤ Q) must contain the index of the lowest common ancestor of Vertex v_i and w_i.

Examples

Input

3 3
5 7
8 11
3 9


Output

2
1
3


Input

100000 2
1 2
3 4


Output

1
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#define FI first
#define SE second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))

vector<lint> head;

constexpr lint INF = 1E9 + 100;
void build_head(lint N)
{
    lint now = 1, sz = 1;
    while (now < INF)
    {
        head.push_back(now);
        if (now > INF or sz > INF) break;
        now += sz;
        sz *= N;
    }
}

lint N;
int find_par(int now)
{
    auto ptr = prev(upper_bound(ALL(head), now));
    auto prvptr = prev(ptr);
    return (now - *ptr) / N + *prvptr;
}

int main()
{
    cin >> N;
    if (N != 2 and N != 1) build_head(N);
    int Q;
    cin >> Q;
    REP(_, Q)
    {
        lint v, w;
        cin >> v >> w;
        if (N == 1)
        {
            printf("%lld\n", min(v, w));
        }
        else if (N != 2)
        {
            while (v != w)
            {
                if (v > w) v = find_par(v);
                else w = find_par(w);
            }
            printf("%lld\n", v);
        }
        else
        {
            while(v != w)
            {
                if (v > w) v /= 2;
                else w /= 2;
            }
            printf("%lld\n", v);
        }
    }
    return 0;
}