Sumo wrestling champion Yamamotoyama (IANS News Photo) 
Entertainment

Japanese Sumo wrestler to enter 'Bigg Boss 5'

The 272-kg wrestling champion Yamamotoyama will make a guest appearance in the reality show.

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NEW DELHI: After The Great Khali entertained the audiences last year in the "Bigg Boss" house, this year Japanese Sumo wrestling champion Yamamotoyama a.k.a. Yama will make a guest appearance in the reality show.

The 601-pound or 272-kg wrestler will add zing to the finale week. However, certain changes will be made in the house to accommodate the six feet four inches tall Yama.

The dining area will be converted into a bedroom-cum-eating area for the wrestler so that he has enough space to do his daily chores. He will be provided with special placards with pictorial representations that will help him to communicate his needs to the housemates.

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