DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
const int N = (1e6);
const int MAX = 7 * (1e7);
const int MOD = (1e9) + 1;
const long long INF = (1e17) + 789;
const double pi = acos(0.0);
long long a[N], t[4 * N], p[4 * N], sum[4 * N];
inline void tree(int v, int l, int r) {
  if (l == r) {
    t[v] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  tree(v * 2, l, mid);
  tree(v * 2 + 1, mid + 1, r);
}
inline void push(int v, int l, int r) {
  if (p[v] != 0) {
    sum[v] += (r - l + 1) * p[v];
    if (l != r) {
      p[v * 2] += p[v];
      p[v * 2 + 1] += p[v];
    }
    if (t[v]) {
      t[v * 2] = t[v * 2 + 1] = t[v];
    }
    p[v] = 0;
  }
}
inline void upd(int v, int l, int r, int ql, int qr, int val) {
  push(v, l, r);
  if (qr < l || r < ql) return;
  if (ql <= l && r <= qr && t[v]) {
    p[v] += abs(t[v] - val);
    t[v] = val;
    push(v, l, r);
    return;
  }
  push(v, l, r);
  int mid = (l + r) / 2;
  upd(v * 2, l, mid, ql, qr, val);
  upd(v * 2 + 1, mid + 1, r, ql, qr, val);
  t[v] = (t[v * 2] == t[v * 2 + 1] ? t[v * 2] : 0);
  sum[v] = sum[v * 2] + sum[v * 2 + 1];
}
long long get(int v, int l, int r, int ql, int qr) {
  push(v, l, r);
  if (qr < l || r < ql) return 0ll;
  if (ql <= l && r <= qr) return sum[v];
  int mid = (l + r) / 2;
  return get(v * 2, l, mid, ql, qr) + get(v * 2 + 1, mid + 1, r, ql, qr);
}
inline void start(int n) {
  for (int i = 1; i <= n + 1; i++) {
    a[i] = i;
  }
  for (int i = 1; i <= 4 * n + 1; i++) {
    p[i] = 0ll;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int n, m;
  cin >> n >> m;
  start(n);
  tree(1, 1, n);
  for (int i = 1; i <= m; i++) {
    char ch;
    cin >> ch;
    if (ch == '1') {
      int ql, qr, val;
      cin >> ql >> qr >> val;
      upd(1, 1, n, ql, qr, val);
    } else {
      int ql, qr;
      cin >> ql >> qr;
      cout << get(1, 1, n, ql, qr) << "\n";
    }
  }
  return false;
}
