Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(long long a) {
  write(a);
  puts("");
}
inline void wri(long long a) {
  write(a);
  putchar(' ');
}
long long rnd() {
  unsigned long long ans = 0;
  for (int i = (int)(0); i <= (int)(4); i++) ans = ans << 15 ^ rand();
  return ans % ((unsigned long long)1 << 63);
}
const int N = 4100, M = N << 1, B = 257;
vector<int> v[N], V[M];
vector<pair<long long, long long> > ans;
vector<vector<int> > T[M];
int tot, nodecnt, a[N], n, q, rt[N], an[70000];
int ask(int nod, int l, int r) {
  if (l == r) return l;
  l = lower_bound(V[nod].begin(), V[nod].end(), l) - V[nod].begin();
  r = upper_bound(V[nod].begin(), V[nod].end(), r) - V[nod].begin() - 1;
  return l > r ? 0 : T[nod][l][r];
}
int mer(int x, int y) {
  if (!x || !y) return x | y;
  if (x == y) return x;
  ++tot;
  ans.push_back(make_pair(x, y));
  return tot;
}
void dfs(vector<int> v, int l, int r, int &rt) {
  rt = ++nodecnt;
  int n = v.size(), ls, rs;
  if (n > 1) {
    vector<int> vl, vr;
    int mid = (l + r) >> 1;
    for (auto i : v)
      if (a[i] <= mid)
        vl.push_back(i);
      else
        vr.push_back(i);
    dfs(vl, l, mid, ls);
    dfs(vr, mid + 1, r, rs);
  }
  T[rt].resize(n);
  for (int i = (int)(0); i <= (int)(n - 1); i++) {
    T[rt][i].resize(n);
    for (int j = (int)(i); j <= (int)(n - 1); j++) {
      T[rt][i][j] = mer(ask(ls, v[i], v[j]), ask(rs, v[i], v[j]));
    }
  }
  V[rt] = v;
}
int main() {
  n = read();
  q = read();
  for (int i = (int)(1); i <= (int)(n); i++) a[i] = read();
  tot = n;
  for (int i = 0; i <= n; i += B) {
    vector<int> v;
    for (int j = (int)(1); j <= (int)(n); j++)
      if (a[j] >= i && a[j] < i + B) {
        v.push_back(j);
      }
    dfs(v, i, i + B, rt[i / B]);
  }
  for (int i = (int)(1); i <= (int)(q); i++) {
    int l = read(), r = read(), ans = 0;
    for (int i = (int)(0); i <= (int)(n / B); i++)
      ans = mer(ans, ask(rt[i], l, r));
    an[i] = ans;
  }
  writeln(tot);
  for (auto i : ans) {
    wri(i.first);
    writeln(i.second);
  }
  for (int i = (int)(1); i <= (int)(q); i++) wri(an[i]);
}