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;
template <class T>
inline void checkmin(T &a, const T &b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, const T &b) {
  if (b > a) a = b;
}
const int inf = 1e9 + 10;
const int maxn = 2e5 + 10;
bool same[maxn * 4];
long long cnt[maxn * 4], upd[maxn * 4], tree[maxn * 4][2], add[maxn * 4];
int n, m;
long long absx(long long x) { return x < 0 ? -x : x; }
void addval(int x, int l, int r, int pos, long long val) {
  if (l == r) {
    cnt[x] = 1;
    tree[x][0] = val;
    tree[x][1] = 0;
    upd[x] = 0;
    same[x] = 1;
    add[x] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    addval(x << 1, l, mid, pos, val);
  else
    addval(x << 1 | 1, mid + 1, r, pos, val);
  tree[x][1] = tree[x << 1][1] + tree[x << 1 | 1][1];
  cnt[x] = cnt[x << 1] + cnt[x << 1 | 1];
  same[x] = 0;
  upd[x] = 0;
  tree[x][0] = 0;
}
void push(int x, int l, int r) {
  if (!upd[x]) return;
  add[x << 1] += add[x];
  add[x << 1 | 1] += add[x];
  tree[x << 1][1] += add[x] * cnt[x << 1];
  tree[x << 1 | 1][1] += add[x] * cnt[x << 1 | 1];
  tree[x << 1][0] = tree[x << 1 | 1][0] = upd[x];
  upd[x << 1] = upd[x << 1 | 1] = upd[x];
  upd[x] = 0;
  add[x] = 0;
}
long long update(int x, int l, int r, int cl, int cr, long long val) {
  if (r < cl || l > cr) return tree[x][1];
  if (cl <= l && r <= cr) {
    if (!same[x]) {
      int mid = (l + r) >> 1;
      long long v1 = update(x << 1, l, mid, cl, cr, val);
      long long v2 = update(x << 1 | 1, mid + 1, r, cl, cr, val);
      same[x] = 1;
      tree[x][0] = val;
      tree[x][1] = v1 + v2;
      return tree[x][1];
    } else {
      push(x, l, r);
      upd[x] = val;
      same[x] = 1;
      add[x] = absx(tree[x][0] - val);
      tree[x][1] += cnt[x] * add[x];
      tree[x][0] = val;
      return tree[x][1];
    }
  }
  push(x, l, r);
  same[x] = 0;
  int mid = (l + r) >> 1;
  long long v1 = update(x << 1, l, mid, cl, cr, val);
  long long v2 = update(x << 1 | 1, mid + 1, r, cl, cr, val);
  tree[x][1] = v1 + v2;
  return tree[x][1];
}
long long getval(int x, int l, int r, int cl, int cr) {
  if (r < cl || l > cr) return 0ll;
  if (cl <= l && r <= cr) return tree[x][1];
  push(x, l, r);
  int mid = (l + r) >> 1;
  long long v1 = getval(x << 1, l, mid, cl, cr);
  long long v2 = getval(x << 1 | 1, mid + 1, r, cl, cr);
  return v1 + v2;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) addval(1, 0, n - 1, i, i + 1);
  for (int test = 0; test < m; test++) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    l--;
    r--;
    if (type == 1) {
      long long x;
      scanf("%lld", &x);
      update(1, 0, n - 1, l, r, x);
    } else
      printf("%lld\n", getval(1, 0, n - 1, l, r));
  }
  return 0;
}
