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.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
const int nax = 3000 * 1007;
int n, q;
int tab[nax];
ll los[nax];
int m;
pii cons[nax];
ll xo[nax];
map<pii, vi> pozy;
map<pii, vll> hasze;
unordered_map<ll, int> mapa;
int wyn[nax];
void start(int a, int b) {
  if (a < b) {
    start(a, (a + b) >> 1);
    start((a + b + 2) >> 1, b);
  }
  pii p = {a, b};
  for (int i = 1; i <= n; i++)
    if (tab[i] >= a && tab[i] <= b) pozy[p].push_back(i);
  ll s = 0;
  hasze[p].push_back(0);
  for (int i : pozy[p]) {
    s ^= los[i];
    hasze[p].push_back(s);
  }
}
int lacz(int a, int b) {
  if (!a) return b;
  if (!b) return a;
  m++;
  cons[m] = {a, b};
  xo[m] = xo[a] ^ xo[b];
  mapa[xo[m]] = m;
  return m;
}
ll daj(int a, int b, int lim) {
  vi &wek = pozy[{a, b}];
  int x = upper_bound(wek.begin(), wek.end(), lim) - wek.begin();
  return hasze[{a, b}][x];
}
int rek(int a, int b, int dol, int gor) {
  ll ha = daj(dol, gor, b) ^ daj(dol, gor, a - 1);
  if (mapa.count(ha)) return mapa[ha];
  return lacz(rek(a, b, dol, (dol + gor) >> 1),
              rek(a, b, (dol + gor + 2) >> 1, gor));
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &tab[i]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 60; j++) los[i] = (los[i] << 1) ^ (rand() & 1);
    xo[i] = los[i];
    mapa[xo[i]] = i;
  }
  m = n;
  mapa[0] = 0;
  start(1, n);
  for (int i = 1; i <= q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    wyn[i] = rek(a, b, 1, n);
  }
  printf("%d\n", m);
  for (int i = n + 1; i <= m; i++)
    printf("%d %d\n", cons[i].first, cons[i].second);
  for (int i = 1; i <= q; i++) printf("%d ", wyn[i]);
  printf("\n");
  return 0;
}
