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;
const int MAXN = 100000 + 50;
struct node {
  int l, r, c, ctag, sz;
  long long ans, atag;
} tr[MAXN * 4];
void upp(int p) {
  if (tr[(p << 1)].c == tr[((p << 1) | 1)].c)
    tr[p].c = tr[(p << 1)].c;
  else
    tr[p].c = -1;
  tr[p].ans = tr[(p << 1)].ans + tr[((p << 1) | 1)].ans;
}
void dwn(int p) {
  if (tr[p].l == tr[p].r) return;
  int ctag = tr[p].ctag;
  long long atag = tr[p].atag;
  tr[p].atag = tr[p].ctag = 0;
  if (ctag) {
    tr[(p << 1)].c = tr[((p << 1) | 1)].c = ctag;
    tr[(p << 1)].ctag = tr[((p << 1) | 1)].ctag = ctag;
  }
  if (atag) {
    tr[(p << 1)].atag += atag;
    tr[((p << 1) | 1)].atag += atag;
    tr[(p << 1)].ans += tr[(p << 1)].sz * atag;
    tr[((p << 1) | 1)].ans += tr[((p << 1) | 1)].sz * atag;
  }
}
void build(int p, int l, int r) {
  tr[p].l = l;
  tr[p].r = r;
  tr[p].c = -1;
  tr[p].sz = r - l + 1;
  if (l == r) {
    tr[p].c = l;
    return;
  }
  int mid = (l + r) >> 1;
  build((p << 1), l, mid);
  build(((p << 1) | 1), mid + 1, r);
}
void change(int p, int x, int y, int v) {
  dwn(p);
  int l = tr[p].l, r = tr[p].r;
  if (l == x && y == r && tr[p].c != -1) {
    long long t = abs(tr[p].c - v);
    tr[p].ans += tr[p].sz * t;
    tr[p].atag += t;
    tr[p].c = tr[p].ctag = v;
    return;
  }
  int mid = (r + l) >> 1;
  if (y <= mid)
    change((p << 1), x, y, v);
  else if (mid < x)
    change(((p << 1) | 1), x, y, v);
  else {
    change((p << 1), x, mid, v);
    change(((p << 1) | 1), mid + 1, y, v);
  }
  upp(p);
}
long long qrysum(int p, int x, int y) {
  dwn(p);
  int l = tr[p].l, r = tr[p].r;
  if (l == x && y == r) return tr[p].ans;
  int mid = (r + l) >> 1;
  if (y <= mid)
    return qrysum((p << 1), x, y);
  else if (mid < x)
    return qrysum(((p << 1) | 1), x, y);
  else
    return qrysum((p << 1), x, mid) + qrysum(((p << 1) | 1), mid + 1, y);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  for (int i = 1, op, x, y, z; i <= m; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) scanf("%d", &z), change(1, x, y, z);
    if (op == 2) printf("%lld\n", qrysum(1, x, y));
  }
  return 0;
}
