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You are given an infinite periodic array *a*0,<=*a*1,<=...,<=*a**n*<=-<=1,<=... with the period of length *n*. Formally, . A periodic subarray (*l*,<=*s*) (0<=≤<=*l*<=&lt;<=*n*, 1<=≤<=*s*<=&lt;<=*n*) of array *a* is an infinite periodic array with a period of length *s* that is a subsegment of array *a*, starting with position *l*.

A periodic subarray (*l*,<=*s*) is superior, if when attaching it to the array *a*, starting from index *l*, any element of the subarray is larger than or equal to the corresponding element of array *a*. An example of attaching is given on the figure (top — infinite array *a*, bottom — its periodic subarray (*l*,<=*s*)):

Find the number of distinct pairs (*l*,<=*s*), corresponding to the superior periodic arrays.
The first line contains number *n* (1<=≤<=*n*<=≤<=2·105). The second line contains *n* numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=106), separated by a space.
Print a single integer — the sought number of pairs.
In the first sample the superior subarrays are (0, 1) and (3, 2).

Subarray (0, 1) is superior, as *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">0</sub>, *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">0</sub>, ...

Subarray (3, 2) is superior *a*<sub class="lower-index">3</sub> ≥ *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">0</sub>, *a*<sub class="lower-index">3</sub> ≥ *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">0</sub> ≥ *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub> ≥ *a*<sub class="lower-index">3</sub>, ...

In the third sample any pair of (*l*, *s*) corresponds to a superior subarray as all the elements of an array are distinct.