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>
using namespace std;
using ll = long long;
int const N = 100 * 1000 + 16;
int n, q;
ll t[N];
ll z[N];
int c[N];
int a[N];
ll v[N];
int ql, qr;
void build(int i, int l, int r) {
  if (r - l + 1 <= 717) {
    for (int j = l; j <= r; ++j) a[j] = j;
  } else {
    int m = (l + r) >> 1;
    build(2 * i, l, m);
    build(2 * i + 1, m + 1, r);
  }
}
inline void push(int i, int l, int r) {
  if (!z[i] || r - l + 1 <= 717) return;
  if (c[i]) {
    c[2 * i] = c[i];
    c[2 * i + 1] = c[i];
  }
  int m = (l + r) >> 1;
  t[2 * i] += z[i] * (m - l + 1);
  t[2 * i + 1] += z[i] * (r - m);
  z[2 * i] += z[i];
  z[2 * i + 1] += z[i];
  z[i] = 0;
}
void update(int i, int l, int r, int x) {
  push(i, l, r);
  if (c[i] == x) {
    return;
  } else if (ql <= l && r <= qr && c[i]) {
    ll v = abs(x - c[i]);
    c[i] = x;
    t[i] += v * (r - l + 1);
    z[i] += v;
  } else if (r - l + 1 <= 717) {
    int co = c[i];
    c[i] = x;
    for (int j = l; j <= r; ++j) {
      a[j] = co ?: a[j];
      if (ql <= j && j <= qr) {
        int d = abs(a[j] - x);
        a[j] = x;
        v[j] += d;
        t[i] += d;
      } else {
        c[i] *= (a[j] == x);
      }
    }
  } else {
    int m = (l + r) >> 1;
    if (ql <= m) update(2 * i, l, m, x);
    if (m < qr) update(2 * i + 1, m + 1, r, x);
    int lf = c[2 * i];
    int rg = c[2 * i + 1];
    c[i] = (lf && lf == rg) ? lf : 0;
    t[i] = t[2 * i] + t[2 * i + 1];
  }
}
void update(int l, int r, int x) {
  ql = l, qr = r;
  update(1, 1, n, x);
}
ll query(int i, int l, int r) {
  push(i, l, r);
  if (ql <= l && r <= qr) {
    return t[i];
  } else if (r - l + 1 <= 717) {
    l = max(l, ql);
    r = min(r, qr);
    ll x = z[i] * (r - l + 1);
    for (int j = l; j <= r; ++j) x += v[j];
    return x;
  } else {
    int m = (l + r) >> 1;
    ll x = (ql <= m) ? query(2 * i, l, m) : 0;
    x += (m < qr) ? query(2 * i + 1, m + 1, r) : 0;
    return x;
  }
}
ll query(int l, int r) {
  ql = l, qr = r;
  return query(1, 1, n);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) a[i] = i;
  build(1, 1, n);
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      int x;
      cin >> x;
      update(l, r, x);
    } else {
      cout << query(l, r) << '\n';
    }
  }
  cout.flush();
}
