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;
const int mod = 998244353, B = 260, N = 4200, T = 20, M = 2300000, Q = 100000;
int n, q, a[N], ans[T][B][B], tot, out[Q], p[B << 1][B][B], cnt;
pair<int, int> bel[M];
vector<int> ind[T], tmp;
inline int addMod(int a, int b) { return (a += b) >= mod ? a - mod : a; }
inline long long quickpow(long long base, long long pw) {
  long long ret = 1;
  while (pw) {
    if (pw & 1) ret = ret * base % mod;
    base = base * base % mod, pw >>= 1;
  }
  return ret;
}
template <class T>
inline void read(T &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
  return;
}
inline int merge(int x, int y) {
  if (!x || !y) return x + y;
  ++tot, bel[tot] = make_pair(x, y);
  return tot;
}
int solve(int lt, int rt, vector<int> &b) {
  int ret = ++cnt;
  if (lt == rt) {
    p[ret][0][0] = b[0];
    return ret;
  }
  int mid = (lt + rt) >> 1;
  vector<int> l, r;
  for (auto &v : b) (a[v] <= mid ? l : r).push_back(v);
  int ls = solve(lt, mid, l), rs = solve(mid + 1, rt, r);
  auto u = l.begin(), v = r.begin();
  while (u != l.end() || v != r.end()) {
    int lf = u - l.begin(), rf = v - r.begin();
    auto uu = u, vv = v;
    while (uu != l.end() || vv != r.end()) {
      ++(uu == l.end() || (vv != r.end() && *vv < *uu) ? vv : uu);
      int ld = uu - l.begin(), rd = vv - r.begin();
      if (!ld)
        p[ret][rf][rd - 1] = p[rs][rf][rd - 1];
      else if (!rd)
        p[ret][lf][ld - 1] = p[ls][lf][ld - 1];
      else
        p[ret][lf + rf][ld + rd - 1] =
            merge(p[ls][lf][ld - 1], p[rs][rf][rd - 1]);
    }
    ++(u == l.end() || (v != r.end() && *v < *u) ? v : u);
  }
  return ret;
}
int main() {
  int x, y;
  read(n), read(q), tot = n;
  for (register int i = 1; i <= n; ++i) {
    read(a[i]), --a[i];
    ind[a[i] >> 8].push_back(i);
  }
  for (register int i = 0; i <= (n - 1) >> 8; ++i) {
    cnt = 0;
    int w = solve(i << 8, min((i + 1) << 8, n) - 1, ind[i]);
    for (register int j = 0, sz = ind[i].size(); j < sz; ++j)
      for (register int k = j; k < sz; ++k) ans[i][j][k] = p[w][j][k];
  }
  for (register int i = 1; i <= q; ++i) {
    read(x), read(y);
    tmp.clear();
    for (register int j = 0; j <= (n - 1) >> 8; ++j) {
      int l = lower_bound(ind[j].begin(), ind[j].end(), x) - ind[j].begin();
      int r = upper_bound(ind[j].begin(), ind[j].end(), y) - ind[j].begin() - 1;
      if (l > r) continue;
      tmp.push_back(ans[j][l][r]);
    }
    int lst = 0;
    for (auto &v : tmp) lst = merge(lst, v);
    out[i] = lst;
  }
  cout << tot << '\n';
  for (register int i = n + 1; i <= tot; ++i)
    cout << bel[i].first << ' ' << bel[i].second << '\n';
  for (register int i = 1; i <= q; ++i) cout << out[i] << ' ';
  cout << endl;
  return 0;
}